3. Tests for Parallelograms
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2 &Given:&&AC≅CF & &&AB≅CD≅BE & &&DF≅DE &Prove:&&BE∥CD Proof.
Congruent segments have the same length, so AC=CF. According to the Segment Addition Postulate, we can write both sides as a sum, AB+BC=CD+DF. Segments AB and CD are congruent, so we can subtract AB=CD to get BC=DF. Segments DF and DE are also congruent, so we can replace DF with DE to get BC=DE. This means that segments BC and DE are congruent. Since it is given that segments CD and BE are also congruent, Theorem 6.9 guarantees that BCDE is a parallelogram. Opposite sides of a parallelogram are parallel, so BE∥CD.
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